Laplace inverse of given question

Question

$$\mathcal{L}^{-1}\left(\frac{2 s - 1}{s^{4} + s^{2} + 1}\right)=~~?$$

I have done the $~\dfrac{2s}{s^4+s^2+1}~$.

But what to do with the $~\dfrac{1}{s^4+s^2+1}~$ ?

I don't get any idea after decomposition.

Answer 1

$$s^4+s^2+1=(s^2+1)^2-s^2=(s^2-s+1)(s^2+s+1)$$

$$\dfrac{2s}{s^4+s^2+1}=\dfrac{(s^2+s+1)-(s^2-s+1)}{(s^2-s+1)(s^2+s+1)}=?$$

Use $$L(e^{at}\sin bt)=\dfrac b{(s-a)^2+b^2}$$

Answer 2

emathhelp gives,

$$\mathcal{L}^{-1}\left(\frac{1}{s^{4} + s^{2} + 1}\right)=\frac{\sqrt{3}}{3} \left(- e^{t} \cos{\left(\frac{\sqrt{3} t}{2} + \frac{\pi}{6} \right)} + \sin{\left(\frac{\sqrt{3} t}{2} + \frac{\pi}{3} \right)}\right) e^{- \frac{t}{2}}$$and $$\mathcal{L}^{-1}\left(\frac{2 s}{s^{4} + s^{2} + 1}\right)=\frac{4}{3} \sqrt{3} \sin{\left(\frac{\sqrt{3} t}{2} \right)} \sinh{\left(\frac{t}{2} \right)}$$ Hence $$\mathcal{L}^{-1}\left(\frac{2 s - 1}{s^{4} + s^{2} + 1}\right)=\left(e^{t} \sin{\left(\frac{\sqrt{3} t}{2} + \frac{\pi}{6} \right)} - \frac{5}{6} \sqrt{3} \sin{\left(\frac{\sqrt{3} t}{2} \right)} - \frac{1}{2} \cos{\left(\frac{\sqrt{3} t}{2} \right)}\right) e^{- \frac{t}{2}}$$